时间戳

首页 > 解决方案 > 为什么我的程序对于大于 2^29 的数字会失败？

c++ - 为什么我的程序对于大于 2^29 的数字会失败？

问题描述

我制作了以下程序，它采用长度为 2^i 的列表“a”（初始化为全 1），并将其包含的所有数字相加。当 i 至少为 30 时，它会返回一个无意义的答案。我不明白为什么，我对所有东西都使用了 long，在我的机器上 long 的大小是 8 字节 = 64 位，所以我会说它必须能够容纳高达 2^(8 * 8)/ 的整数2.

// FOR NOW ONLY WORKS WITH N A POWER OF 2
#include "cuda_runtime.h"
#include "device_launch_parameters.h"

#include <stdio.h>
#include <iostream>
#include <chrono>

/* 
Parallel reduce helper function. When run 
with n/2 threads, adds a[n - 1 - i] to a[i]
for i = 0, ..., n - 1.
*/
__global__ void reduce(long* a, long n)
{
    long i = threadIdx.x + blockDim.x * blockIdx.x;
    long stride = gridDim.x * blockDim.x;

    for (long j = i; j < n/2; j += stride)
    {
        a[j] += a[n - 1 - j];
    }
}

/* 
For an array a of length n, puts the sum of all elements in a[0]
*/
void parallelReduce(long* a, long n)
{
    // Get some information about the GPU
    cudaDeviceProp prop;
    cudaGetDeviceProperties(&prop, 0);
    int multiProcessors = prop.multiProcessorCount;

    // Repeatedly use the helper function reduce
    while (n > 1) {
        int threadsPerBlock = 256;
        int numberOfBlocks = 32 * multiProcessors;
        reduce << <numberOfBlocks, threadsPerBlock >> > (a, n);
        cudaDeviceSynchronize();
        n = (n + n % 2) / 2; // Rounds n/2 up.
    }
}

int main()
{
    // Initialize vector with N 1's.
    long N = 2 << 28;
    size_t size = N * sizeof(long);
    long* h_a;
    cudaMallocHost(&h_a, size);

    for (long i = 0; i < N; i++) {
        h_a[i] = 1;
    }

    // Copy to device (can be done asynchronically to hide transfer time, but 
    // that messes up the timing of the kernel).
    long* d_a;
    cudaMalloc(&d_a, size);
    cudaMemcpy(d_a, h_a, size, cudaMemcpyHostToDevice);

    // Calculate the sum sequentially and time it.
    auto tic = std::chrono::high_resolution_clock::now();
    long hostSolution = 0;
    for (long i = 0; i < N; i++)
    {
        hostSolution += h_a[i];
    }
    auto toc = std::chrono::high_resolution_clock::now();
    int duration = std::chrono::duration_cast<std::chrono::milliseconds>(toc - tic).count();

    std::cout << "The sequential function says the answer is " << hostSolution << " this took " << duration
        << " ms." << std::endl;

    // Kernel computation
    tic = std::chrono::high_resolution_clock::now();
    parallelReduce(d_a, N);
    toc = std::chrono::high_resolution_clock::now();
    int parallelDuration = std::chrono::duration_cast<std::chrono::milliseconds>(toc - tic).count();

    // Copy result back to host
    long solution;
    cudaMemcpy(&solution, &d_a[0], sizeof(long), cudaMemcpyDeviceToHost);

    // Print the parallel result and speed up:
    std::cout << "The parallel function says the answer is " << solution << " this took " << parallelDuration
        << " ms." << std::endl;

    std::cout << "This means we have achieved a speed up of " << duration / parallelDuration << std::endl;
}

标签： c++typescuda

解决方案

我们可以，您只需要正确的类型。编译并执行以下代码。

#include <iostream>
#include <limits>

int main() {
    std::cout << "Max int value: " << std::numeric_limits<int>::max() << '\n';
    std::cout << "Max long value: " << std::numeric_limits<long>::max() << '\n';
    std::cout << "Max long long value: " << std::numeric_limits<long long>::max() << '\n';
}

您的输出取决于您的 ide/architecture/compiler flags 和其他东西，对我来说是以下内容。

Max int value: 2147483647
Max long value: 2147483647
Max long long value: 9223372036854775807

至于为什么'当 i 至少 30 岁时，它返回一个无意义的答案'，溢出是 UB，你无法传达编译器在这种情况下会做什么。

推荐阅读